Answer
a) $\frac{t}{t^{2}+1}+C$
b) $-\frac{1}{\sqrt {t^{2}+1}}+C$
Work Step by Step
a)
$t$ = $tanθ$
$dt$ = $sec^{2}θdθ$
$\int{\frac{dt}{(t^{2}+1)^{\frac{3}{2}}}}$ = $\int{\frac{sec^{2}θ}{(tan^{2}θ+1)^{\frac{3}{2}}}}dθ$ = $\int{cosθ}dθ$ = $sinθ+C$
$t$ = $tanθ$
$sinθ$ = $\frac{t}{t^{2}+1}$
$\int{\frac{dt}{(t^{2}+1)^{\frac{3}{2}}}}$ = $sinθ+C$ = $\frac{t}{t^{2}+1}+C$
b)
$u$ = $t^{1}+1$
$du$ = $2tdt$
$\int{\frac{tdt}{(t^{2}+1)^{\frac{3}{2}}}}$ = $\frac{1}{2}\int{u^{-\frac{3}{2}}}du$ = $-u^{-\frac{1}{2}}+C$ = $-\frac{1}{\sqrt {t^{2}+1}}+C$
